Question

If A = [a_ij]2×2, where a_ij = $\left\{\begin{array}{l}i+j,\mathrm{if}i\ne j\\ i^2-2j,\mathrm{if}i=j\end{array}\right.$, then A^-1 = ____________________.

Answer 1

Given:
A = [a_ij]_2×2, where a_ij = $\left\{\begin{array}{l}i+j,\mathrm{if}i\ne j\\ i^2-2j,\mathrm{if}i=j\end{array}\right.$


$$\begin{gathered} a_{ij}=\left\{\begin{array}{l}i+j,\mathrm{if}i\ne j\\ i^2-2j,\mathrm{if}i=j\end{array}\right. \\ {a}_{11}={\left(1\right)}^2-2\left(1\right)=-1 \\ {a}_{12}=1+2=3 \\ {a}_{21}=2+1=3 \\ {a}_{22}={\left(2\right)}^2-2\left(2\right)=0 \\ \mathrm{Thus}, \\ A=\left[\begin{array}{cc}-1& 3\\ 3& 0\end{array}\right] \\ \mathrm{Therefore}, \\ {A}^{-1}=\frac{1}{0-9}\left[\begin{array}{cc}0& -3\\ -3& -1\end{array}\right] \\ =-\frac{1}{9}\left[\begin{array}{cc}0& -3\\ -3& -1\end{array}\right] \\ =\frac{1}{9}\left[\begin{array}{cc}0& 3\\ 3& 1\end{array}\right] \end{gathered}$$


Hence, $A^{-1}=\underline{\frac{1}{9}\left[\begin{array}{cc}0& 3\\ 3& 1\end{array}\right]}.$